CBSE Class 10 Mathematics Standard sample paper 2023-24: question 25(a)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. Each removed portion is a sector of radius $r=14$ cm. Its angle is the angle of the triangle at that vertex.
   $$\text{Total area removed}=\frac{\angle A}{360^{\circ }}\pi r^2+\frac{\angle B}{360^{\circ }}\pi r^2+\frac{\angle C}{360^{\circ }}\pi r^2$$

   ½ mark

2. Take $\pi r^2$ common: $=\frac{\angle A+\angle B+\angle C}{360^{\circ }}\pi r^2=\frac{180^{\circ }}{360^{\circ }}\pi r^2$ [angle sum property of a triangle].

   ½ mark

3. Substitute the values: $=\frac{180}{360}\times \frac{22}{7}\times (14{)}^2$.

   ½ mark

4. $=\frac{1}{2}\times \frac{22}{7}\times 196=308{\text{cm}}^2$

   So the total area removed is $308{\text{cm}}^2$.

   ½ mark

Answer: 308 cm²

Revise the chapter

• Chapter: Areas Related to Circles (notes, formulas, where marks are lost and every past question)
• Topic: Area of a Sector of a Circle